Create and Share a File from Internal Storage

My goal is to create a XML file on internal storage and then send it through the share Intent.

I'm able to create a XML file using this code

FileOutputStream outputStream = context.openFileOutput(fileName, Context.MODE_WORLD_READABLE);
PrintStream printStream = new PrintStream(outputStream);
String xml = this.writeXml(); // get XML here
printStream.println(xml);
printStream.close();

I'm stuck trying to retrieve a Uri to the output file in order to share it. I first tried to access the file by converting the file to a Uri

File outFile = context.getFileStreamPath(fileName);
return Uri.fromFile(outFile);

This returns file:///data/data/com.my.package/files/myfile.xml but I cannot appear to attach this to an email, upload, etc.

If I manually check the file length, it's proper and shows there is a reasonable file size.

Next I created a content provider and tried to reference the file and it isn't a valid handle to the file. The ContentProvider doesn't ever seem to be called a any point.

Uri uri = Uri.parse("content://" + CachedFileProvider.AUTHORITY + "/" + fileName);
return uri;

This returns content://com.my.package.provider/myfile.xml but I check the file and it's zero length.

How do I access files properly? Do I need to create the file with the content provider? If so, how?

Update

Here is the code I'm using to share. If I select Gmail, it does show as an attachment but when I send it gives an error Couldn't show attachment and the email that arrives has no attachment.

public void onClick(View view) {
    Log.d(TAG, "onClick " + view.getId());

    switch (view.getId()) {
        case R.id.share_cancel:
            setResult(RESULT_CANCELED, getIntent());
            finish();
            break;

        case R.id.share_share:

            MyXml xml = new MyXml();
            Uri uri;
            try {
                uri = xml.writeXmlToFile(getApplicationContext(), "myfile.xml");
                //uri is  "file:///data/data/com.my.package/files/myfile.xml"
                Log.d(TAG, "Share URI: " + uri.toString() + " path: " + uri.getPath());

                File f = new File(uri.getPath());
                Log.d(TAG, "File length: " + f.length());
                // shows a valid file size

                Intent shareIntent = new Intent();
                shareIntent.setAction(Intent.ACTION_SEND);
                shareIntent.putExtra(Intent.EXTRA_STREAM, uri);
                shareIntent.setType("text/plain");
                startActivity(Intent.createChooser(shareIntent, "Share"));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }

            break;
    }
}

I noticed that there is an Exception thrown here from inside createChooser(...), but I can't figure out why it's thrown.

E/ActivityThread(572): Activity com.android.internal.app.ChooserActivity has leaked IntentReceiver com.android.internal.app.ResolverActivity$1@4148d658 that was originally registered here. Are you missing a call to unregisterReceiver()?

I've researched this error and can't find anything obvious. Both of these links suggest that I need to unregister a receiver.

• ChooserActivity has leaked IntentReceiver
• Why does Intent.createChooser() need a BroadcastReceiver and how to implement?

I have a receiver setup, but it's for an AlarmManager that is set elsewhere and doesn't require the app to register / unregister.

Code for openFile(...)

In case it's needed, here is the content provider I've created.

public ParcelFileDescriptor openFile(Uri uri, String mode) throws FileNotFoundException {
    String fileLocation = getContext().getCacheDir() + "/" + uri.getLastPathSegment();

    ParcelFileDescriptor pfd = ParcelFileDescriptor.open(new File(fileLocation), ParcelFileDescriptor.MODE_READ_ONLY);
    return pfd;
}

Solution 1:

It is possible to expose a file stored in your apps private directory via a ContentProvider. Here is some example code I made showing how to create a content provider that can do this.

Manifest

<manifest xmlns:android="http://schemas.android.com/apk/res/android"
  package="com.example.providertest"
  android:versionCode="1"
  android:versionName="1.0">

  <uses-sdk android:minSdkVersion="11" android:targetSdkVersion="15" />

  <application android:label="@string/app_name"
    android:icon="@drawable/ic_launcher"
    android:theme="@style/AppTheme">

    <activity
        android:name=".MainActivity"
        android:label="@string/app_name">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

    <provider
        android:name="MyProvider"
        android:authorities="com.example.prov"
        android:exported="true"
        />        
  </application>
</manifest>

In your ContentProvider override openFile to return the ParcelFileDescriptor

@Override
public ParcelFileDescriptor openFile(Uri uri, String mode) throws FileNotFoundException {       
     File cacheDir = getContext().getCacheDir();
     File privateFile = new File(cacheDir, "file.xml");

     return ParcelFileDescriptor.open(privateFile, ParcelFileDescriptor.MODE_READ_ONLY);
}

Make sure you have copied your xml file to the cache directory

    private void copyFileToInternal() {
    try {
        InputStream is = getAssets().open("file.xml");

        File cacheDir = getCacheDir();
        File outFile = new File(cacheDir, "file.xml");

        OutputStream os = new FileOutputStream(outFile.getAbsolutePath());

        byte[] buff = new byte[1024];
        int len;
        while ((len = is.read(buff)) > 0) {
            os.write(buff, 0, len);
        }
        os.flush();
        os.close();
        is.close();

    } catch (IOException e) {
        e.printStackTrace(); // TODO: should close streams properly here
    }
}

Now any other apps should be able to get an InputStream for your private file by using the content uri (content://com.example.prov/myfile.xml)

For a simple test, call the content provider from a seperate app similar to the following

    private class MyTask extends AsyncTask<String, Integer, String> {

    @Override
    protected String doInBackground(String... params) {

        Uri uri = Uri.parse("content://com.example.prov/myfile.xml");
        InputStream is = null;          
        StringBuilder result = new StringBuilder();
        try {
            is = getApplicationContext().getContentResolver().openInputStream(uri);
            BufferedReader r = new BufferedReader(new InputStreamReader(is));
            String line;
            while ((line = r.readLine()) != null) {
                result.append(line);
            }               
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try { if (is != null) is.close(); } catch (IOException e) { }
        }

        return result.toString();
    }

    @Override
    protected void onPostExecute(String result) {
        Toast.makeText(CallerActivity.this, result, Toast.LENGTH_LONG).show();
        super.onPostExecute(result);
    }
}

Solution 2:

So Rob's answer is correct I assume but I did it a bit differently. As far as I understand, with the setting in in provider:

android:exported="true"

you are giving public access to all your files?! Anyway, a way to give only access to some files is to define file path permissions in the following way:

<provider
    android:authorities="com.your.app.package"
    android:name="android.support.v4.content.FileProvider"
    android:exported="false"
    android:grantUriPermissions="true">
    <meta-data
        android:name="android.support.FILE_PROVIDER_PATHS"
        android:resource="@xml/file_paths" />
</provider>

and then in your XML directory you define file_paths.xml file as follows:

<paths xmlns:android="http://schemas.android.com/apk/res/android">
    <files-path path="/" name="allfiles" />
    <files-path path="tmp/" name="tmp" />
</paths>

now, the "allfiles" gives the same kind of public permission I guess as the option android:exported="true" but you don't really want that I guess so to define a subdirectory is the next line. Then all you have to do is store the files you want to share, there in that dir.

Next what you have to do is, as also Rob says, obtain a URI for this file. This is how I did it:

Uri contentUri = FileProvider.getUriForFile(context, "com.your.app.package", sharedFile);

Then, when I have this URI, I had to attach to it permissions for other app to use it. I was using or sending this file URI to camera app. Anyway this is the way how I got the other app package info and granted permissions to the URI:

PackageManager packageManager = getPackageManager();
List<ResolveInfo> list = packageManager.queryIntentActivities(cameraIntent, PackageManager.MATCH_DEFAULT_ONLY);
if (list.size() < 1) {
    return;
}
String packageName = list.get(0).activityInfo.packageName;
grantUriPermission(packageName, sharedFileUri, Intent.FLAG_GRANT_WRITE_URI_PERMISSION | Intent.FLAG_GRANT_READ_URI_PERMISSION);

ClipData clipData = ClipData.newRawUri("CAMFILE", sharedFileUri);
cameraIntent.setClipData(clipData);
cameraIntent.setFlags(Intent.FLAG_GRANT_WRITE_URI_PERMISSION);
cameraIntent.setFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
startActivityForResult(cameraIntent, GET_FROM_CAMERA);

I left the code for camera as I did not want to take some other example I did not work on. But this way you see that you can attach permissions to URI itself.

The camera's thing is that I can set it via ClipData and then additionally set permissions. I guess in your case you only need FLAG_GRANT_READ_URI_PERMISSION as you are attaching a file to an email.

Here is the link to help on FileProvider as I based all of my post on the info I found there. Had some trouble finding a package info for camera app though.

Hope it helps.

Solution 3:

If you need to permission other apps to see your app's private files (for Share, or otherwise) you might be able to save some time and just use v4 compat library's FileProvider

Solution 4:

None of the above answers helped. My problem was the point of passing intent extras but I'll walk you through all the steps to share a file.

Step 1: Create a Content Provider This will make the file accessible to whichever app you want to share with. Paste the following in the Manifest.xml file inside the <application></applicatio> tags

            <provider
                android:name="androidx.core.content.FileProvider"
                android:authorities="{YOUR_PACKAGE_NAME}.fileprovider"
                android:exported="false"
                android:grantUriPermissions="true">
                <meta-data
                    android:name="android.support.FILE_PROVIDER_PATHS"
                    android:resource="@xml/provider_paths" />
            </provider>

Step 2: Define paths accessible by the content provider Do this by creating a file called provider_paths.xml (or a name of your choice) under res/xml. Put the following code in the file

    <?xml version="1.0" encoding="utf-8"?>
    <paths>
        <external-path
            name="external"
            path="." />
        <external-files-path
            name="external_files"
            path="." />
        <cache-path
            name="cache"
            path="." />
        <external-cache-path
            name="external_cache"
            path="." />
        <files-path
            name="files"
            path="." />
    </paths>

Step 3: Create the Intent to share the file

Intent intentShareFile = new Intent(Intent.ACTION_SEND);
Uri uri = FileProvider.getUriForFile(getApplicationContext(), getPackageName() + ".fileprovider", fileToShare);
            
intentShareFile.setDataAndType(uri, URLConnection.guessContentTypeFromName(fileToShare.getName()));
//Allow sharing apps to read the file Uri
intentShareFile.addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
//Pass the file Uri instead of the path
intentShareFile.putExtra(Intent.EXTRA_STREAM,
                    uri);
startActivity(Intent.createChooser(intentShareFile, "Share File"));